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# Lesson 20: Derivatives and the Shapes of Curves

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Using the Mean Value Theorem, we can show the a function is increasing on an interval when its derivative is positive on the interval. Changes in the sign of the derivative detect local extrema. We also can use the second derivative to detect concavity and inflection points. This means that the first and second derivative can be used to classify critical points as local maxima or minima

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### Lesson 20: Derivatives and the Shapes of Curves

1. 1. Section 4.3 Derivatives and the Shapes of Curves V63.0121, Calculus I March 25-26, 2009 . . Image credit: cobalt123 . . . . . .
2. 2. Outline Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Derivative Test Concavity Deﬁnitions Testing for Concavity The Second Derivative Test . . . . . .
3. 3. The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b), then f is decreasing on (a, b). . . . . . .
4. 4. The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f′ > 0 on (a, b), then f is increasing on (a, b). If f′ < 0 on (a, b), then f is decreasing on (a, b). Proof. It works the same as the last theorem. Pick two points x and y in (a, b) with x < y. We must show f(x) < f(y). By MVT there exists a point c in (x, y) such that f(y) − f(x) = f′ (c) > 0. y−x So f(y) − f(x) = f′ (c)(y − x) > 0. . . . . . .
5. 5. Finding intervals of monotonicity I Example Find the intervals of monotonicity of f(x) = 2x − 5. . . . . . .
6. 6. Finding intervals of monotonicity I Example Find the intervals of monotonicity of f(x) = 2x − 5. Solution f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞). . . . . . .
7. 7. Finding intervals of monotonicity I Example Find the intervals of monotonicity of f(x) = 2x − 5. Solution f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞). Example Describe the monotonicity of f(x) = arctan(x). . . . . . .
8. 8. Finding intervals of monotonicity I Example Find the intervals of monotonicity of f(x) = 2x − 5. Solution f′ (x) = 2 is always positive, so f is increasing on (−∞, ∞). Example Describe the monotonicity of f(x) = arctan(x). Solution 1 Since f′ (x) = is always positive, f(x) is always increasing. 1 + x2 . . . . . .
9. 9. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. . . . . . .
10. 10. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. . . . . . .
11. 11. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. 0 . . . . . . .
12. 12. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. ↘ ↗ 0 . f . . . . . . . . .
13. 13. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. ↘ ↗ 0 . f . . . So f is decreasing on (−∞, 0) and increasing on (0, ∞). . . . . . .
14. 14. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. ↘ ↗ 0 . f . . . So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) . . . . . .
15. 15. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). . . . . . .
16. 16. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − × + . .. . . −1/3 x 0 . − + . . 0 .. .x+4 5 − . 4/5 . . . . . .
17. 17. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − × + . .. . . −1/3 x 0 . − + . . 0 .. .x+4 5 − . 4/5 . ′ (x) × f .. 0 .. − 0 . . 4/5 . (x) f . . . . . .
18. 18. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − × + . .. . . −1/3 x 0 . − + . . 0 .. .x+4 5 − . 4/5 . ′ (x) 0−× f + + . .. . . . . ↗ − ↘. ↗ 0 . . 4/5 . . . (x) f . . . . . .
19. 19. The First Derivative Test Theorem (The First Derivative Test) Let f be continuous on [a, b] and c a critical point of f in (a, b). If f′ (x) > 0 on (a, c) and f′ (x) < 0 on (c, b), then c is a local maximum. If f′ (x) < 0 on (a, c) and f′ (x) > 0 on (c, b), then c is a local minimum. If f′ (x) has the same sign on (a, c) and (c, b), then c is not a local extremum. . . . . . .
20. 20. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. ↘ ↗ 0 . f . . . So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) . . . . . .
21. 21. Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solution f′ (x) = 2x, which is positive when x > 0 and negative when x is. We can draw a number line: .′ − f + . . 0 .. ↘ ↗ 0 . f . . . m . in So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) . . . . . .
22. 22. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − × + . .. . . −1/3 x 0 . − + . . 0 .. .x+4 5 − . 4/5 . ′ (x) 0−× f + + . .. . . . . ↗ − ↘. ↗ 0 . . 4/5 . . . (x) f . . . . . .
23. 23. Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solution f′ (x) = 2 x−1/3 (x + 2) + x2/3 = 1 x−1/3 (5x + 4) 3 3 The critical points are 0 and and −4/5. − × + . .. . . −1/3 x 0 . − + . . 0 .. .x+4 5 − . 4/5 . ′ (x) 0−× f + + . .. . . . . ↗ − ↘. ↗ 0 . . 4/5 . . . (x) f m . ax m . in . . . . . .
24. 24. Outline Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Derivative Test Concavity Deﬁnitions Testing for Concavity The Second Derivative Test . . . . . .
25. 25. Deﬁnition The graph of f is called concave up on and interval I if it lies above all its tangents on I. The graph of f is called concave down on I if it lies below all its tangents on I. . . concave up concave down We sometimes say a concave up graph “holds water” and a concave down graph “spills water”. . . . . . .
26. 26. Deﬁnition A point P on a curve y = f(x) is called an inﬂection point if f is continuous there and the curve changes from concave upward to concave downward at P (or vice versa). . concave up i .nﬂection point . . . concave down . . . . . .
27. 27. Theorem (Concavity Test) If f′′ (x) > 0 for all x in I, then the graph of f is concave upward on I If f′′ (x) < 0 for all x in I, then the graph of f is concave downward on I . . . . . .
28. 28. Theorem (Concavity Test) If f′′ (x) > 0 for all x in I, then the graph of f is concave upward on I If f′′ (x) < 0 for all x in I, then the graph of f is concave downward on I Proof. Suppose f′′ (x) > 0 on I. This means f′ is increasing on I. Let a and x be in I. The tangent line through (a, f(a)) is the graph of L(x) = f(a) + f′ (a)(x − a) f(x) − f(a) = f′ (b). So By MVT, there exists a b between a and x with x−a f(x) = f(a) + f′ (b)(x − a) ≥ f(a) + f′ (a)(x − a) = L(x) . . . . . .
29. 29. Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . . . . . . .
30. 30. Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . Solution We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2. . . . . . .
31. 31. Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . Solution We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2. This is negative when x < −1/3, positive when x > −1/3, and 0 when x = −1/3 . . . . . .
32. 32. Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . Solution We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2. This is negative when x < −1/3, positive when x > −1/3, and 0 when x = −1/3 So f is concave down on (−∞, −1/3), concave up on (1/3, ∞), and has an inﬂection point at (−1/3, 2/27) . . . . . .
33. 33. Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). . . . . . .
34. 34. Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f′′ (x) = −x (5x − 2) =x x 9 9 9 . . . . . .
35. 35. Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f′′ (x) = −x (5x − 2) =x x 9 9 9 x−4/3 is always positive, so the concavity is determined by the 5x − 2 factor . . . . . .
36. 36. Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solution 10 −1/3 4 −4/3 2 −4/3 f′′ (x) = −x (5x − 2) =x x 9 9 9 x−4/3 is always positive, so the concavity is determined by the 5x − 2 factor So f is concave down on (−∞, 2/5), concave up on (2/5, ∞), and has an inﬂection point when x = 2/5 . . . . . .
37. 37. The Second Derivative Test Theorem (The Second Derivative Test) Let f, f′ , and f′′ be continuous on [a, b]. Let c be be a point in (a, b) with f′ (c) = 0. If f′′ (c) < 0, then f(c) is a local maximum. If f′′ (c) > 0, then f(c) is a local minimum. If f′′ (c) = 0, the second derivative test is inconclusive (this does not mean c is neither; we just don’t know yet). . . . . . .
38. 38. Example Find the local extrema of f(x) = x3 + x2 . . . . . . .
39. 39. Example Find the local extrema of f(x) = x3 + x2 . Solution f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3. . . . . . .
40. 40. Example Find the local extrema of f(x) = x3 + x2 . Solution f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3. Remember f′′ (x) = 6x + 2 . . . . . .
41. 41. Example Find the local extrema of f(x) = x3 + x2 . Solution f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3. Remember f′′ (x) = 6x + 2 Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum. . . . . . .
42. 42. Example Find the local extrema of f(x) = x3 + x2 . Solution f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3. Remember f′′ (x) = 6x + 2 Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum. Since f′′ (0) = 2 > 0, 0 is a local minimum. . . . . . .
43. 43. Example Find the local extrema of f(x) = x2/3 (x + 2) . . . . . .
44. 44. Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = (5x + 4) which is zero when x = −4/5 x 3 . . . . . .
45. 45. Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = (5x + 4) which is zero when x = −4/5 x 3 10 −4/3 Remember f′′ (x) = (5x − 2), which is negative when x 9 x=− 4/5 . . . . . .
46. 46. Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = (5x + 4) which is zero when x = −4/5 x 3 10 −4/3 Remember f′′ (x) = (5x − 2), which is negative when x 9 x=− 4/5 So x = −4/5 is a local maximum. . . . . . .
47. 47. Example Find the local extrema of f(x) = x2/3 (x + 2) Solution 1 −1/3 Remember f′ (x) = (5x + 4) which is zero when x = −4/5 x 3 10 −4/3 Remember f′′ (x) = (5x − 2), which is negative when x 9 x=− 4/5 So x = −4/5 is a local maximum. Notice the Second Derivative Test doesn’t catch the local minimum x = 0 since f is not differentiable there. . . . . . .
48. 48. Graph Graph of f(x) = x2/3 (x + 2): y . . −4/5, 1.03413) ( . . ( . 2/5, 1.30292) . . x . . −2, 0) ( ( . 0, 0) . . . . . .
49. 49. When the second derivative is zero At inﬂection points c, if f′ is differentiable at c, then f′′ (c) = 0 Is it necessarily true, though? . . . . . .
50. 50. When the second derivative is zero At inﬂection points c, if f′ is differentiable at c, then f′′ (c) = 0 Is it necessarily true, though? Consider these examples: g(x) = −x4 f(x) = x4 h(x) = x3 . . . . . .
51. 51. When the second derivative is zero At inﬂection points c, if f′ is differentiable at c, then f′′ (c) = 0 Is it necessarily true, though? Consider these examples: g(x) = −x4 f(x) = x4 h(x) = x3 All of them have f′′ (0) = 0. But the ﬁrst has a local min at 0, the second has a local max at 0, and the third has an inﬂection point at 0. This is why we say 2DT has nothing to say when f′′ (c) = 0. . . . . . .
52. 52. Summary Concepts: Mean Value Theorem, monotonicity, concavity Facts: derivatives can detect monotonicity and concavity Techniques for drawing curves: the Increasing/Decreasing Test and the Concavity Test Techniques for ﬁnding extrema: the First Derivative Test and the Second Derivative Test . . . . . .
53. 53. Summary Concepts: Mean Value Theorem, monotonicity, concavity Facts: derivatives can detect monotonicity and concavity Techniques for drawing curves: the Increasing/Decreasing Test and the Concavity Test Techniques for ﬁnding extrema: the First Derivative Test and the Second Derivative Test Next week: Graphing functions . . . . . .